STP16NF06 datasheet confusion

[The Magic Rabbit]

I know there are better FETs for what I'm doing, but I have about 40 of these doing nothing so would rather use than waste.

I have a small 20W 12VDC heater that I want to control from a PIC, which will run a timer and thermostat. The output is 5V.

How do I work out how much power the FET would dissipate? I thought there would be a graph showing how RDS changes with VGS, but there isn't. The FET is a STP16NF06

Thanks

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[michaeliv]

It's not in there most likely because it's not guaranteed. You can try and measure it yourself.

[Paul Moir]

Given that the Vgs threshold (the part where the MOSFET just starts conducting) is 4v max, you'll likely need more than 5v out of the pic to drive it reasonably on.  Note that the I/O output of your PIC might be somewhat less than 5v, often Vcc-0.6 or 0.7 v.  However, a single small transistor should be more than enough to get you some decent Vgs. 

I assume you're doing digital full on/full off control here.

(** some PICs have an open drain output which can be exploited to give you a little higher voltage.  RA4 on the 16F84A times to mind.)

[Phoenix]

http://www.st.com/web/en/resource/technical/document/datasheet/CD00002501.pdf

Figure 5 gives you the information you need to estimate the conduction loss. For a given Vgs and Id you find the Vds.

[The Magic Rabbit]

OK, so assuming it was 4.5V from the PIC, that would still be be over my 2A IDS requirement with VDS at 12V... am I reading that correctly?

... however there's no figure for RDS at those values, so it's "suck it and see" essentially?

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[biffster]

Figure 6 plots VGS against ID at a constant VDS. You can calculate RGS from VDS/ID.

Beware though, Figure 8 tells you that RDS(ON) varies significantly with temperature even if VGS is held constant. The temperature of the FET will depend on how much power is being dissipated by it and how good the thermal conductivity to the surroundings are i.e. if you're dissipating a lot of power and you have no heat sink it will get hot and RDS(ON) will go up a lot so your current might be limited.

I'd try breadboarding it and see what you get.

[The Magic Rabbit]

What motor? It's driving a heater.

I've breadboarded it and ran it for a couple of hours. The heater got monstrously hot but the FET wasn't even noticeably warm. Maybe 1 or 2 degrees above ambient. :-)



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[Phoenix]

OK, so assuming it was 4.5V from the PIC, that would still be be over my 2A IDS requirement with VDS at 12V... am I reading that correctly?

... however there's no figure for RDS at those values, so it's "suck it and see" essentially?

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Each curve in Figure 5 is for a different Vgs. You're applying 4.5V so we will need to visually interpolate between the 4V and 5V curves, sort of.
You're trying to conduct 2A so you go up along the Y axis to 2A; the very large scale is hard to accurate read 2A, it's right down near the bottom.
Go right until 2A intercepts the 4.5V curve (somewhere between the 4V and 5V curve) and go down to read Vds is about 700mV.

This is the process, but because you're using such a low gate voltage the MOSFET may not actually be turning fully on, then your Vds will be a higher and your Id will be smaller.

FYI - you can think of RDSon as a useful approximation of the "fully on" (also called the linear or ohmic region) region of the curve in Figure 5. The caveat is that the MOSFET is fully switched on.
https://en.wikipedia.org/wiki/MOSFET#Modes_of_operation
In your circuit the MOSFET may not be operating in this region because Vgs is so low, which means RDSon doesn't apply anyway.

[The Magic Rabbit]

Thank you! This makes a lot more sense now. :-)

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